runtime error: member access within misaligned address 0xbebebebebebebebe for type 'struct MyListNod

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Line 18: Char 26: runtime error: member access within misaligned address 0xbebebebebebebebe for type 'struct MyListNode', which requires 8 byte alignment (solution.c)

不知道LeetCode出了啥问题
我的提交代码

struct List {
    int count;
    struct MyListNode *head;
    struct MyListNode *tail;
};
struct MyListNode {
    int val;
    struct MyListNode *next;
};
void add(struct List *list, int val) {
    struct MyListNode *node = (struct MyListNode *)malloc(sizeof(struct MyListNode *));
    node->val = val;
    if (list->head == NULL) {
        list->head = node;
        list->tail = node;
        list->count = 0;
    } else {
        list->tail->next = node;
        list->tail = node;
    }
    list->count ++;
}
struct Stack {
    struct StackNode *top;
};
struct StackNode {
    struct TreeNode *node;
    struct StackNode *bottom;
};
void push(struct Stack *stack, struct TreeNode *node) {
    struct StackNode *newStackNode = (struct StackNode *)malloc(sizeof(struct StackNode));
    newStackNode->bottom = stack->top;
    newStackNode->node = node;
    stack->top = newStackNode;
}
struct TreeNode * pop(struct Stack *stack) {
    struct StackNode * top  = stack->top;
    stack->top = top->bottom;
    return top->node;
}
int isEmpty(struct Stack * stack) {
    if (stack->top == NULL) {
        return 1;
    } else {
        return 0;
    }
}
/**
 * 前序遍历流程
 * 1. 根节点入栈
 * 2. 从栈中取出一个节点
 * 3. 打印节点数据
 * 4. 如果右节点不为空，入栈
 * 5. 如果左节点不为空，入栈
 * 6. 如果栈不为空，跳转到2
 * @param root
 * @param returnSize
 * @return
 */
int* preorderTraversal(struct TreeNode* root, int* returnSize) {
    if (root == NULL) {
        *returnSize = 0;
        return (int*)malloc(sizeof(int) * 0);
    }
    struct Stack *stack = (struct Stack *)malloc(sizeof(struct Stack));
    struct List *list = (struct List *)malloc(sizeof(struct List));
    push(stack, root);
    while (!isEmpty(stack)) {
        root = pop(stack);
        if (root == NULL){
            break;
        }
        int val = root->val;
        add(list, val);
        if (root->right != NULL) {
            push(stack, root->right);
        }
        if (root->left != NULL) {
            push(stack, root->left);
        }
    }
    int *ret = (int *)malloc(sizeof(int) * list->count);
    struct MyListNode * it = list->head;
    for(int i=0; i<list->count; i++) {
        ret[i] = it->val;
        it = it->next;
    }
    *returnSize = list->count;
    return ret;
}

在本地可以跑起来的代码

#include <stdio.h>
#include <stdlib.h>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/**
 * Return an array of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};
struct List {
    int count;
    struct MyListNode *head;
    struct MyListNode *tail;
};
struct MyListNode {
    int val;
    struct MyListNode *next;
};
void add(struct List *list, int val) {
    struct MyListNode *node = (struct MyListNode *)malloc(sizeof(struct MyListNode *));
    node->val = val;
    if (list->head == NULL) {
        list->head = node;
        list->tail = node;
        list->count = 0;
    } else {
        list->tail->next = node;
        list->tail = node;
    }
    list->count ++;
}
struct Stack {
    struct StackNode *top;
};
struct StackNode {
    struct TreeNode *node;
    struct StackNode *bottom;
};
void push(struct Stack *stack, struct TreeNode *node) {
    struct StackNode *newStackNode = (struct StackNode *)malloc(sizeof(struct StackNode));
    newStackNode->bottom = stack->top;
    newStackNode->node = node;
    stack->top = newStackNode;
}
struct TreeNode * pop(struct Stack *stack) {
    struct StackNode * top  = stack->top;
    stack->top = top->bottom;
    return top->node;
}
int isEmpty(struct Stack * stack) {
    if (stack->top == NULL) {
        return 1;
    } else {
        return 0;
    }
}
/**
 * 前序遍历流程
 * 1. 根节点入栈
 * 2. 从栈中取出一个节点
 * 3. 打印节点数据
 * 4. 如果右节点不为空，入栈
 * 5. 如果左节点不为空，入栈
 * 6. 如果栈不为空，跳转到2
 * @param root
 * @param returnSize
 * @return
 */
int* preorderTraversal(struct TreeNode* root, int* returnSize) {
    if (root == NULL) {
        *returnSize = 0;
        return (int*)malloc(sizeof(int) * 0);
    }
    struct Stack *stack = (struct Stack *)malloc(sizeof(struct Stack));
    struct List *list = (struct List *)malloc(sizeof(struct List));
    push(stack, root);
    while (!isEmpty(stack)) {
        root = pop(stack);
        if (root == NULL){
            break;
        }
        int val = root->val;
        add(list, val);
        if (root->right != NULL) {
            push(stack, root->right);
        }
        if (root->left != NULL) {
            push(stack, root->left);
        }
    }
    int *ret = (int *)malloc(sizeof(int) * list->count);
    struct MyListNode * it = list->head;
    for(int i=0; i<list->count; i++) {
        ret[i] = it->val;
        it = it->next;
    }
    *returnSize = list->count;
    return ret;
}
struct TreeNode * newTreeNode(int val) {
    struct TreeNode * node = (struct TreeNode *)malloc(sizeof(struct TreeNode));
    node->val = val;
    return node;
}
int main() {
    struct TreeNode *root = newTreeNode(1);
    struct TreeNode *right = newTreeNode(2);
    struct TreeNode *left = newTreeNode(3);
    root->right = right;
    right->left = left;
    int count = 0;
    int *ret = preorderTraversal(root, &count);
    printf("count: %d\n", count);
    for(int i=0;i < count; i++) {
        printf("%d ", ret[i]);
    }
    printf("\n");
}